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Proof step 2 from proof del-del-one (previous step)

⊦ (∅ ++ a) -- b = ∅ ++ a, a = b

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s;% ⊦ a ∈ ∅ ++ b ↔ a = b; (s)% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)